现代结构分析方法 (07－08年度第一学期)
第8讲
正倒空间对应性
r N
= rrhk*l
⊥ ( hkl )
rrh*kl = 1 d hkl
倒易点阵、正点阵与衍射
c 入射束
相机长度
(hkl)
S/λ= k
S0/λ= k0 ghkl
000
ab c
ba 正交点阵沿c轴投影d100
b a
Rhkl
d110
d100
indexing using PDF files
R1= R2= 16.7mm, R1^R2=90º. Lλ=3.0 nm*mm.
R2
R3
R1
a) Choose three spots such as R3=h3k3l3, R1=h1k1l1, R2=h2k2l2. b) di=Lλ/Ri, d1=Lλ/R1= d2= 0.180nmÆ{200}, d3=Lλ/R3=0.127nmÆ{220}. c) h3k3l3= h1k1l1 + h2k2l2: {200}+{200}={220}Æ(200)+(020)=(220), (200)^(020)=90º d) [uvw] = R1×R2= R1×R3= R3×R2=[004] =[001]
(hkl)
S/λ= k
θ
S0/λ= k0 L
ghkl
(hkl)
000
Rhkl ’
Rhkl
[-101]
(hkl)
7.2 Plotting and indexing of single crystal spot patterns
S/λ= k
相机长度
S0/λ= k0
(010)
ghkl
(101)
000
Rhkl
NSY= 1
NL= 1
SY: 1-CUBIC; 2-TETRA; 3-ORTH; 4-HEX;
R2
5-MONO; 6-TRIC
LT: 1-F; 2-I; 3-C; 4-B; 5-A; 6-P; 7-R;
R1= 16.7 mm, R2 = R3= 27.5 mm, R1^R2=107.5º. Lλ=48.226 A*mm.
F：The 4 lattice points per unit cell are located at 000, 0.5 0.5 0, 0 0.5 0.5, 0.5 0 0.5. Then Fhkl= f {exp[-2πi (h 0 +k 0 +l 0)] + exp[-2πi (h 0 +k 1/2 +l 1/2)] + exp[-2πi (h 1/2 +k 1/2 +l 0)] + exp[-2πi (h 1/2 +k 0 +l 1/2)]}= f {1 + cosπ(k+l) + cosπ(l+h) + cosπ(h+k)}.
120.00 2.051 2.051
2 1 1 0 -1 1 -1 -1 1 1 1.000 1.155
70.53 3.349 3.349
3 1 0 0 0 -2 0 0 0 -2 1.000 1.414
90.00 2.900 2.900
4 3 3 2 2 -2 0 1 1 -3 1.173 1.541
70.53 3.349 3.349
3 1 0 0 0 -2 0 0 0 -2 1.000 1.414
90.00 2.900 2.900
4 3 3 2 2 -2 0 1 1 -3 1.173 1.541
90.00 2.051 1.749
5 2 2 1 2 -2 0 0 2 -4 1.581 1.581
000
Rhkl
6、 衍射和对称性：反射条件
φg＝Σ Fn exp[-2πiK.rn] 衍射在倒易阵点上发生
F(θ) = Σfj(θ) exp[-2πi(k – k0).rj] = Σfj(θ) exp[-2πi (h xj +k yj +l zj)], Ihkl ∝ ⏐Fhkl⏐2, 倒易阵点的衍射强度受结构因子调制 （2） 点阵类型的影响：
7.2 Plotting and indexing of
(hkl)
single crystal spot patterns
相机长度
S/λ= k
S0/λ= k0 ghkl
000
For a known substance but unknown orientation, a table of interplanar spacings d is needed.
111 R2
R1 111
220 R3
109.5º
R1=R2=14.4mm, R1^R2=109.5º. Lλ=48.226 A*mm[1.10]
K U V W H1 K1 L1 H2 K2 L2 R2/R1 R3/R1 FAI
d1
d2
1 1 1 1 0 2 -2 -2 0 2 1.000 1.000
a) Choose three spots such as h3k3l3, h1k1l1, h2k2l2. b) Measure the d values, and thus determine the indices.
Rhkl
h3k3l3
h2k2l2
h1k1l1
R2 R1
[uvw]
c) By trial and error a consistent set of indices is chosen such that h3k3l3= h1k1l1 + h2k2l2.
R1= R2= 16.7mm, R1^R2=90º. Lλ=48.226 A*mm.
K U V W H1 K1 L1 H2 K2 L2 R2/R1 R3/R1 FAI
d1
d2
1 1 1 1 0 2 -2 -2 0 2 1.000 1.000
120.00 2.051 2.051
2 1 1 0 -1 1 -1 -1 1 1 1.000 1.155
108.43 2.051 1.297
6 2 1 1 1 -1 -1 0 2 -2 1.633 1.915
90.00 3.349 2.051
PARAMETERS A= 5.8000 B= 5.8000 C= 5.8000 Å
R3 R1
AF= 90.000 BT= 90.000 GM= 90.000
NUVW= 3
90.00 2.051 1.749
5 2 2 1 2 -2 0 0 2 -4 1.581 1.581
Hale Waihona Puke 108.43 2.051 1.297
6 2 1 1 1 -1 -1 0 2 -2 1.633 1.915
90.00 3.349 2.051
PARAMETERS A= 5.8000
AF= 90.000
R2 R1
[uvw]
a) Choose three spots such as R3=h3k3l3, R1=h1k1l1, R2=h2k2l2. b) di=Lλ/Ri, d1=Lλ/R1= d2= 0.208nmÆ{111}, R3=Lλ/R3=3/16.6=0.180nmÆ{200}. c) h3k3l3= h1k1l1 + h2k2l2: {111}+{111}={200}Æ(111)+(1-1-1)=(200), (111)^(1-1-1)=109.5º d) [uvw] = R1×R2= R1×R3= R3×R2=[01-1]
indexing using PDF files
R1= 16.7 mm, R2 = R3= 27.5 mm, R1^R2=107.5º. Lλ=3.0 nm*mm.
R3 R1 R2
a) Choose three spots such as R3=h3k3l3, R1=h1k1l1, R2=h2k2l2. b) di=Lλ/Ri, d1=Lλ/R1= 0.180nmÆ{200}, R2 = R3=Lλ/R2=0.109nmÆ{311}. c) h3k3l3= h1k1l1 + h2k2l2: {200}+{311}={311}Æ(200)+(-113)=(113), (200)^(-113)=107.5º d) [uvw] = R1×R2= R1×R3= R3×R2=[0-31]
衍射强度受到结构因子Fn调制 倒易矢量 正空间点阵矢量
φg＝Σ Fn exp[-2πiK.rn]
n1n2n3 第n个单胞的位置
K2 K1
点积为整数时发生衍射Æ晶面方程
K1不指向倒易阵点 K2指向倒易阵点(-1-10)
5，衍射仪原理
相机长度
(hkl)
S/λ= k θ
S0/λ= k0 ghkl
(hkl)
b a d110
[001]
(010) (100) (110)
220
010
110
b* 000 a* 100
3.5 单胞散射结构因子
单胞内所有原子散射波的总和，振幅正比于 F(θ) = Σfj(θ) exp[-2πi(k – k0).rj]， 第j个原子的位置坐标
3.6 完整晶体散射 衍射发生在倒易点阵节点hkl位置
indexing using PDF files
R1= R2= 16.7mm, R1^R2=90º. Lλ=3.0 nm*mm.
R2
R3
R1
a) Choose three spots such as R3=h3k3l3, R1=h1k1l1, R2=h2k2l2. b) di=Lλ/Ri, d1=Lλ/R1= d2= 0.180nmÆ{200}, d3=Lλ/R3=0.127nmÆ{220}. c) h3k3l3= h1k1l1 + h2k2l2: {200}+{200}={220}Æ(200)+(020)=(220), (200)^(020)=90º d) [uvw] = R1×R2= R1×R3= R3×R2=[004] =[001]